Question

The dielectric constant of the interior of a protein is considerably smaller than that of water. How would this difference in dielectric constants affect the strength of an electrostatic interaction between two opposite charges with the same distance between them if the charged groups were located in the interior of the protein rather than on its surface

Answer 1

Answer:

the interaction in the protein is greater than the surface with water

\frac{F_i}{F_s} = \frac{\epsilon_s}{ \epsilon_i} \ > 1

Explanation:

The electric force  for a charge is

          F = $\frac{1}{4\pi \epsilon} \ \frac{q^2}{r^2}$

In the exercise indicate that the charge is q and the distance r is maintained, the test charge is another  

therefore if we use the index i for the dielectric constant ($\epsilon_i$) in the protein

         $F_{i} = \frac{1}{4\pi \epsilon_i} \frac{q^2}{r^2}$  

the electric force in water with dielectric constant ($\epsilon_s$)

           $F_s = \frac{1}{4\pi \epsilon_s} \frac{q^2}{r^2}$

            $\epsilon_i < \epsilon_s$

if we look for the relationship between these forces

          $\frac{F_i}{F_s} = \frac{\epsilon_s}{ \epsilon_i} \ > 1$

therefore the interaction in the protein is greater than the surface with water